Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{q^2 + 3q - 28}{10q + 10} \times \dfrac{-9q - 9}{q^2 - 4q} $
Answer: First factor the quadratic. $a = \dfrac{(q - 4)(q + 7)}{10q + 10} \times \dfrac{-9q - 9}{q^2 - 4q} $ Then factor out any other terms. $a = \dfrac{(q - 4)(q + 7)}{10(q + 1)} \times \dfrac{-9(q + 1)}{q(q - 4)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (q - 4)(q + 7) \times -9(q + 1) } { 10(q + 1) \times q(q - 4) } $ $a = \dfrac{ -9(q - 4)(q + 7)(q + 1)}{ 10q(q + 1)(q - 4)} $ Notice that $(q + 1)$ and $(q - 4)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -9\cancel{(q - 4)}(q + 7)(q + 1)}{ 10q(q + 1)\cancel{(q - 4)}} $ We are dividing by $q - 4$ , so $q - 4 \neq 0$ Therefore, $q \neq 4$ $a = \dfrac{ -9\cancel{(q - 4)}(q + 7)\cancel{(q + 1)}}{ 10q\cancel{(q + 1)}\cancel{(q - 4)}} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $a = \dfrac{-9(q + 7)}{10q} ; \space q \neq 4 ; \space q \neq -1 $